I have a question related to an equation that I once found but now I am unable to figure out what it is exactly used for. It is an equation for variance but I am unable to figure out what sort of variance calculation it is used for. If anyone can kindly help me out with this, I will be really thankful. This is the equation:

$$\text{Unknown thing} = \frac{2 \sigma^2}{n} \Bigg[ \frac{1+(r-1)\rho}{r} - \frac{(m+1)\rho-1}{m} \Bigg].$$


Given the total absence of context or source information, this is quite a tricky reverse-engineering problem, but let me see if I am up to the task. Based on the structure of the equation, it appears to me that it is some kind of variance equation pertaining to sample means of equicorrelated random variables.

Let's start our analysis with a simpler observation. If you have $n$ equicorrelated random variables with correlation $\rho$ then the variance of the sample mean is:

$$\begin{align} \mathbb{V}(\bar{X}_n) &= \mathbb{V} \bigg( \frac{1}{n} \sum_{i=1}^n X_i \bigg) \\[6pt] &= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \mathbb{C}(X_i,X_j) \\[6pt] &= \frac{1}{n^2} \Bigg[ \sum_{i} \mathbb{V}(X_i) + \sum_{i \neq j} \mathbb{C}(X_i,X_j) \Bigg] \\[6pt] &= \frac{1}{n^2} \Bigg[ \sum_{i} \sigma^2 + \sum_{i \neq j} \rho \sigma^2 \Bigg] \\[6pt] &= \frac{\sigma^2}{n} \Bigg[ n + n(n-1) \rho \Bigg] \\[6pt] &= \frac{1 + (n-1) \rho}{n} \cdot \sigma^2. \\[6pt] \end{align}$$

(Note that there is a range restriction $-1/(n-1) \leqslant \rho \leqslant 1$ to ensure that the variance matrix is non-negative definite.) This gives you part of the equation from the first bracketed term. I am unable to see how you would obtain the second bracketed term from any well-known manipulation of equicorrelated random variables. Nevertheless, the equation looks to me like something you might get when conducting some kind of comparison of variances of sample means of equicorrelated random variables.

Hopefully that information gets you some way to identifying where you saw this equation. Without some further context or source information, this is the best I can do to help you identify what it might be talking about.

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  • $\begingroup$ Hello Ben, thank you so much for the response and yes without context it has become difficult to reverse engineer. However, I believe this has to do with the calculation of variance for the difference in difference estimator now that you have hinted at it being a comparison of variances of sample means of correlated random variables. Rho being the autocorrelation but no idea what r and m are then. $\endgroup$ – Fahad Jul 4 at 17:01
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    $\begingroup$ The values $r$ and $m$ are almost certainly sample sizes for two groups, and perhaps $n=r+m$ is the total size (not certain of the last part). I recommend you investigate the variance form that emerges from looking at diff-in-diff for equicorrelated random variables. $\endgroup$ – Ben - Reinstate Monica Jul 4 at 22:44

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